Problem: What is the slope of the line tangent to $f(x) = -x^{2}+3x+5$ at $x = 5$ ?
Solution: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{(-(x+h)^{2}+3(x+h)+5) - (-x^{2}+3x+5)}{h}$ $ = \lim_{h \to 0} \frac{(-(x^{2}+2x h+h^{2})+3(x+h)+5) - (-x^{2}+3x+5)}{h}$ $ = \lim_{h \to 0} \frac{-x^{2}-2(x h)-h^{2}+3x+3h+5+x^{2}-3x-5}{h}$ $ = \lim_{h \to 0} \frac{-2(x h)-h^{2}+3h}{h}$ $ = \lim_{h \to 0} -2x-h+3$ $ = -2x+3$ $ = (-2)(5)+3$ $ = -7$